31
Vertices of a right triangle ABC with $\angle B = 90^\circ$ are $A(3, 4)$, $B(1, 1)$ and $C(-8, 7)$. Find the value of $\tan A$.
Show SolutionHide Solution↓
Sol. $BC = \sqrt{(-8 - 1)^2 + (7 - 1)^2} = \sqrt{117} = 3\sqrt{13}$ (I Mark)
$AB = \sqrt{(3 - 1)^2 + (4 - 1)^2} = \sqrt{13}$ (II Mark)
$\tan A = \frac{BC}{AB} = \frac{3\sqrt{13}}{\sqrt{13}} = 3$ (III Mark)
$AB = \sqrt{(3 - 1)^2 + (4 - 1)^2} = \sqrt{13}$ (II Mark)
$\tan A = \frac{BC}{AB} = \frac{3\sqrt{13}}{\sqrt{13}} = 3$ (III Mark)